Solution:
For odd $n$ it is impossible to create a table like that. Indeed, in such a table the sum of all column sums and row sums would be $1+2+\ldots+2n = n(2n+1)$, and so would be odd. But it would also be twice the sum of all the numbers in the table (each number counted twice - once in the column sum, once in the row sum), and so would have to be even. This contradiction proves that no such table exists.

For any even $n$ such a table exists. If one takes a $2k$ by $2k$ table and fills in all the elements above the main diagonal (running from upper left to lower right corner of the table) with $2$'s, all the elements below the main diagonal with $0$'s, first $k$ elements on the diagonal with $1$'s and the last $k$ elements on the main diagonal with $2$'s, one gets a table that satisfies the conditions. Indeed, the sums of elements in the first $k$ rows (top to bottom) are $4k-1, 4k-3, \ldots, 2k+1$, the next $k$ rows $2k, 2k-2, \ldots, 2$. The sums of elements in the first $k$ columns (left to right) are $1, 3, \ldots, 2k-1$, the next $k$ columns $2k+2, \ldots, 4k$. We see that all the numbers from $1$ to $4k$ appear exactly once, as wanted.